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Math Help - Radius and and Height

  1. #1
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    Radius and and Height

    This is my practice exam question and I have no answer for it nor do I know how to go about this....any help is greatly appreciated.

    An open-topped cylindrical water reservoir has a volume of 320{\pi}m^3.
    The material used for the bottom of the reservoir costs $5/ m^2 and the material used for the vertical portion costs $8/ m^2.
    Determine the radius and the height of the cylinder so that the cost is a minimum.
    They also provide me with these equations:
    Volume = {\pi}r^2h
    surface are = {\pi}r^2 + 2 {\pi}rh
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by lil_cookie View Post
    This is my practice exam question and I have no answer for it nor do I know how to go about this....any help is greatly appreciated.

    An open-topped cylindrical water reservoir has a volume of 320{\pi}m^3.
    The material used for the bottom of the reservoir costs $5/ m^2 and the material used for the vertical portion costs $8/ m^2.
    Determine the radius and the height of the cylinder so that the cost is a minimum.
    They also provide me with these equations:
    Volume = {\pi}r^2h
    surface are = {\pi}r^2 + 2 {\pi}rh
    Restriction:  \pi r^2 h = 320

       h = \frac{320}{\pi r^2}
    Side Area = 2 Pi r H
    Base Area = Pi r^2

    Side Area Cost = 2 \pi r h * 8
    Base Area Cost = \pi r^2 *5

    Total Cost = 5 \pi r^2 + 16 \pi r h

    Plug in the expression for h in the cost function, then derive with respect to r to optimize.

    Let me know if you need further help with this.
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  3. #3
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    could you continue answering it....I want to see how it's done

    Thank You
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  4. #4
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    Quote Originally Posted by lil_cookie View Post
    could you continue answering it....I want to see how it's done

    Thank You
    Please show the working you've attempted in response to the help you've been given so far. Where do you get stuck?
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  5. #5
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    So I need to solve for r in this equation:

    Total Cost = 5{\pi}r^2 + 16{\pi}r\frac{320}{{\pi}r^2}
    Then I move {\pi}r^2 to other side:
     <br />
{\pi}r^2 = 5{\pi}r^2 + 16{\pi}r*320<br />

    Then I'm not sure how to get r alone to solve for it....can someone please show me....my exam is today....Thanks
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  6. #6
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    Quote Originally Posted by lil_cookie View Post
    So I need to solve for r in this equation:

    Total Cost = 5{\pi}r^2 + 16{\pi}r\frac{320}{{\pi}r^2}

    Mr F says: Why are you doing the following??:

    Then I move {\pi}r^2 to other side:
     <br />
{\pi}r^2 = 5{\pi}r^2 + 16{\pi}r*320<br />

    Then I'm not sure how to get r alone to solve for it....can someone please show me....my exam is today....Thanks
    You must have done or seen similar problems!

    Once you have the expression for total cost, you differentiate it with respect to r and then solve dT/dr = 0. Get r, test the nature etc.

    This might be an excellent time to go back and refer to your class notes and/or textbook.
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